This sounded like a trick question at first, but having spent a little time on it, the pattern is somewhat surprising and gives a remarkably high number of results divisible by 3!

The answer I get seems to tend towards

, but it does vary depending on the value of n. (eg, if n=1, then it is 33.33%, as I had expected).

Having worked on this the hard way, let's now try to get a formula.

Mathsy, I agree with your logic, but the question asked for two numbers to be selected at random, not the same number to be selected twice. Hence, I excluded the possibility of x=y. This has the effect of removing 3n from both numerator and denominator. Progressing through the series, up to very high values of n will still tend towards 55.55%, but always slightly less than this and, at very low values of n, significantly less.

A technicality, maybe, but if I am playing poker, I like to know how many aces are in the deck!

Try this:

[((3n)^2 x 5/9) -3n] / [3n x (3n-1)]

***2

Solve for x.

(x-1)³+(x-2)³+(x-3)³+(x-4)³+(x-5)³=0

***2
do you want complex numbers?
Eq=5(x-3)(x^2-6x+15)
so x=3 or x=3+I√6 or x=3-I√6

***2

Well, the answer is

, but my algebra is so rusty that I am struggling to come up with the elegant solution rather than the trial and error version! Give me time and I WILL get there!

Thanks, Krassi.

You were obviously posting your answer while I was still working on mine, so I saw yours after posting mine! I had got as far as multiplying out and simplifying, but I just hadn't got as far as the final factorisation. It used to be so easy when I was 18, as I was doing it all the time, but you don't have much need for this kind of maths as a hotelier, so it takes me a bit longer to get my thinking in gear!!

What a lovely compliment! I wish that my teacher had agreed with you in 1978 (or, even the examination board!!). I was far from the perfect student and my maths was only average at best. But what I have done in life since then is to learn from mistakes, work from first principles wherever possible and always try to understand WHY things are the way they are. It may take a lot longer than remembering formulae, but it is much more sound in the long-term.

Alas, question ***3 is currently beyond me and I really need to dust away a lot more cobwebs before I can suggest a properly worked solution, but I do agree with 1 + 5e being the correct answer, based on observation rather than on algebraic proof.

I wanted to say another thing from drags but the word blocker from the forum didn't allowed me to do it.

***4

How many number of n digits can be made with nonzero digits in which no two consecutive digits are the same?

Excellent, mathsyperson!

ganesh, I am seriously struggling with the proof of ****3. I can work through the logical structure of the 4 given equations and I can see an obvious similarity with the expression to be solved. However, for the life of me, I just cannot get an expression that absolutely links this expression to those funtions of e. The problem is that the numerator's progression is a funtion of (1+2n)^2 rather than x^n, as given in the examples, which is destined to diverge greatly as n increases.

My solution was simply based around common sense, in that the answer HAD to have a relationship to e, otherwise the examples are pointless!. Doing a sum of the first few values of the expression soon led to an answer of 14.5914, which is clearly 1 + 5e. Equally, the progression follows a very similar pattern to the formula for e, so it was clear that it is not necessary to bring in high values of x as their effect is negligable.

The worked example that gives this missing link would be greatly appreciated!